Algebraic identities

This is a difference of two perfect squares: \(49p^2=(7p)^2\) and \(64q^2=(8q)^2\). Hence the identity \(a^2-b^2\) applies. Exam tip: first check whether both terms are perfect squares.

The correct identity is \(p^2-q^2=(p+q)(p-q)\). On multiplying, the middle terms \(-pq\) and \(+pq\) cancel, leaving \(p^2-q^2\). Exam tip: check cancellation of middle terms.

\(9x^2-12xy+4y^2=(3x-2y)^2\). Its middle term is \(-2(3x)(2y)=-12xy\). In option B, the last term is not \(4y^2\). Exam tip: check the middle term using twice the product of the square roots.

In the square of three terms, the mixed term formed by two selected terms is twice their product. Here the relevant terms are 2x and z, so the xz term is

2(2x)(z)=4xz.

The term -6yz comes from the product of -3y and z, so it is not an xz term. Exam tip: identify the required pair of terms and take twice their product.

In \(x^2+6x+9\), the middle term is \(2\cdot x\cdot3=6x\) and the last term is \(3^2=9\), so it equals \((x+3)^2\). \(x^2+6x+8\) fails because its constant term is not 9. Exam tip: check \(a^2+2ab+b^2\).

The constant term (-4k=-28) gives (k=7), and the middle term also becomes (3x). Exam tip: verify using both conditions.

The first is (x^2+6x+9) and the second is (x^2+6x-7). Subtracting gives (9-(-7)=16), so the correct option should be (16).

The first product is (4x^2+20x+9) and the second is (4x^2+20x+25). Exam tip: cancel like terms and compare constants.

The correct identity is \(a^2-b^2=(a+b)(a-b)\). Multiplying gives \(a^2-ab+ab-b^2=a^2-b^2\). In contrast, \((a-b)^2\) contains an extra \(-2ab\) term. Exam tip: look for conjugate binomials with opposite signs.

The sum identity gives (2(5a)^2+2(4b)^2=50a^2+32b^2). Exam tip: notice the cancellation of middle terms.

The expression contains two squares that differ only in the sign of the middle term. Group the terms that are unchanged by writing the expression as \\(A+y)^2-(A-y)^2\\), where \\(A=x+z\\). The identity \\( (A+B)^2-(A-B)^2=4AB\\) says that this difference is four times the product of the common part and the changing term. This avoids expanding every term separately.

Substituting \\(A=x+z\\) and \\(B=y\\) gives \\(4(x+z)y=4y(x+z)\\). Thus the correct simplified form is option A. Option B is exactly half the required value, option C keeps only an incomplete product, and option D does not represent the difference of these two squares. The grouping and identity also provide a quick check on the result.

A three-term square has double products of every pair. Here the (pq), (pr), and (qr) terms are (12pq), (6pr), and (4qr).

This is ( (4x)^2-2\cdot4x\cdot5y+(5y)^2 ). Exam tip: identify a minus square from the negative middle term.

\(81a^2-49b^2=(9a)^2-(7b)^2\). Applying the identity \(x^2-y^2=(x+y)(x-y)\) gives \((9a+7b)(9a-7b)\). Options B and D are perfect squares and would produce a middle term of \(\mp126ab\), which is absent in the given expression. Exam tip: identify the square roots of both terms, then write one sum factor and one difference factor.

The sum of constants is (12+(-5)=7) and the product is (-60). Exam tip: check sum and product separately with mixed signs.

\((p+q)(p-q)\) are conjugate binomials, and their product is \(p^2-q^2\), a difference of two squares. In \((p+q)^2\), the middle term is \(2pq\). Exam tip: look for the same terms with opposite signs.

Use the identity
\((a-b)^2=a^2-2ab+b^2\). Thus,
\((4m-3n)^2=16m^2-24mn+9n^2\). Subtracting
\((16m^2+9n^2)\) cancels the square terms, leaving
\(-24mn\).
\(24mn\) has the wrong sign because the middle term in a square of a difference is negative. Exam tip: in
\((a-b)^2\), the middle term is always
\(-2ab\).

Use the identities \((a+b)^2=a^2+2ab+b^2\) and \((a-b)^2=a^2-2ab+b^2\). Their sum is \(2a^2+2b^2\), since the \(+2ab\) and \(-2ab\) terms cancel. Subtracting \(2a^2+2b^2\) therefore gives \(0\). \(2ab\) is not correct because the middle terms add to zero. Exam tip: remember \((x+y)^2+(x-y)^2=2x^2+2y^2\).

Both expansions have the same (x^2+14x), and constants give (40-48=-8). Exam tip: cancel like terms and subtract constants.

The pair terms in the square are (6ab), (6ac), and (2bc). Exam tip: take the double product of every pair.